Collisions: Elastic, Inelastic and the Coefficient of Restitution

One number decides every head-on collision between two bodies: e, the coefficient of restitution, which is 1 for elastic, 0 for a perfectly inelastic collision where they stick, and anything in between for the real world. Give both masses, both signed velocities and e, and this returns both final velocities, the energy lost, and the momentum check.

Last updated 11 Nov 2026

One-dimensional collision, parameterised by e

e = 1 elastic, e = 0 they stick, in between real

Mass of body 1In kilograms. Both masses may be anything; the formulae are symmetric, so it does not matter which body you call 1 as long as you keep its velocity in the field below it.

Velocity of body 1 before, signedIn m/s, signed along a direction you choose as positive. A head-on collision means the two velocities have opposite signs; a rear-end collision means the same sign with the one behind moving faster. Everything on the page depends on getting these two signs right, and nothing on the page can warn you if you have not.

Mass of body 2In kilograms. Set it very large — a few thousand times body 1 — to model a ball hitting a wall or the floor, which is the commonest use of the coefficient of restitution outside an exam.

Velocity of body 2 before, signedIn m/s, same positive direction. Zero is the standard textbook case, a moving body striking a stationary one.

Coefficient of restitution, eDimensionless, between 0 and 1 inclusive — Edexcel’s Further Mathematics specification lists the inequality 0 ≤ e ≤ 1 as examinable content, and this field enforces it. It is defined as the relative speed of SEPARATION divided by the relative speed of APPROACH. e = 1 is a perfectly elastic collision and conserves kinetic energy; e = 0 is perfectly inelastic and the bodies move off together; in between, some energy is lost and the bodies separate more slowly than they approached. To measure it for a ball dropped onto a hard floor, e is the square root of the rebound height divided by the drop height.

-1.2000m/sExample

2 kg at +6 m/s meeting 3 kg at −2 m/s head-on, with e = 0.5

Momentum conserved, separation speed e times approach speed, and two unknowns

m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂ · v₂ − v₁ = e(u₁ − u₂) · v₁ = [(m₁ − em₂)u₁ + (1 + e)m₂u₂] / (m₁ + m₂)

conservation of momentum: always true, for every e, provided no external force acts during the collision. This is the equation that does not care whether energy is lost, which is why it is the one to write down first and why the page prints the total before and after as a check
e = separation / approach: Newton’s experimental law, and the second equation you need because momentum alone gives one equation for two unknowns. Note it is a ratio of RELATIVE speeds, not of individual ones. Edexcel’s Further Mechanics 1 names it “Newton’s law of restitution” and examines 0 ≤ e ≤ 1
e = 1: elastic. Kinetic energy is conserved, the bodies separate exactly as fast as they approached, and the relative velocity simply reverses. Cambridge 9702 states this as a required outcome without ever naming e: for an elastic collision “the relative speed of approach is equal to the relative speed of separation”
e = 0: perfectly inelastic. The bodies do not separate at all, so they move off together at the velocity of the centre of mass, and the energy lost is the largest any collision at these speeds can lose
what is NOT conserved: kinetic energy, unless e = 1. The missing energy goes into permanent deformation, heating and sound. Total energy is conserved, of course; kinetic energy is not a conserved quantity and treating it as one is the commonest wrong assumption in collision questions

Worked example

2 kg at +6 m/s meeting 3 kg at −2 m/s head-on, with e = 0.5
THE TWO EQUATIONS. Momentum: 2(6) + 3(−2) = +6 kg m/s, so 2v₁ + 3v₂ = 6. Restitution: the approach speed is 6 − (−2) = 8 m/s, so the separation speed is 0.5 × 8 = 4 m/s and v₂ − v₁ = 4. Two equations, two unknowns.
SOLVE. From the second, v₂ = v₁ + 4. Substituting: 2v₁ + 3v₁ + 12 = 6, so 5v₁ = −6 and v₁ = −1.2000 m/s — body 1 has been thrown back the way it came. Then v₂ = +2.8000 m/s, so body 2 has reversed as well.
CHECK THE MOMENTUM. After: 2(−1.2) + 3(2.8) = −2.4 + 8.4 = +6 kg m/s, the same as before. This check costs one line and catches almost every arithmetic slip, because conservation of momentum holds for every value of e and so the total cannot move whatever else you got wrong.
NOW THE ENERGY. Before: ½(2)(36) + ½(3)(4) = 36 + 6 = 42 J. After: ½(2)(1.44) + ½(3)(7.84) = 1.44 + 11.76 = 13.2 J. So 28.8 J — 68.6 per cent — has gone, into deformation and heat. Momentum unchanged, two thirds of the energy destroyed: that is what an inelastic collision is.
AND THE CENTRE OF MASS, which is the elegant check. v_cm = 6/5 = +1.2 m/s before the collision and +1.2 m/s after it, because no external force acted. Every collision between these two bodies, at every value of e from 0 to 1, leaves the centre of mass moving at exactly +1.2 m/s. All e does is decide how fast the bodies move relative to that.

The same two bodies at every e, from elastic to perfectly inelastic

One pair of bodies — 2 kg at +6 m/s meeting 3 kg at −2 m/s — at seven values of e, with g playing no part at all. Three things to read out of it. First, the momentum after is +6 kg m/s in EVERY row, which is why the centre of mass moves at +1.2 m/s whatever e is; the rows differ only in how the bodies move relative to it. Second, the separation speed falls linearly with e, from 8 m/s to zero, because that is the definition of e and not a consequence of anything. Third, the energy lost is largest at e = 0 and zero at e = 1, and the perfectly inelastic row still keeps 3.6 J — the kinetic energy of the combined body moving at +1.2 m/s, which momentum conservation will not let it give up. Only a collision whose total momentum is zero can lose all of its kinetic energy. Notice too where body 1 stops rebounding: somewhere between e = 0.4 and e = 0.2 its final velocity changes sign, so whether it bounces back or is carried along depends on e and not only on the masses.
e	What it means	v₁, m/s	v₂, m/s	Separation speed, m/s	KE after, J	KE lost, J	KE lost, %
1.0	perfectly elastic	-3.6000	+4.4000	8.0000	42.0000	0.0000	0.0%
0.8	very bouncy	-2.6400	+3.7600	6.4000	28.1760	13.8240	32.9%
0.6	a lively ball	-1.6800	+3.1200	4.8000	17.4240	24.5760	58.5%
0.5	the worked example	-1.2000	+2.8000	4.0000	13.2000	28.8000	68.6%
0.4	dull	-0.7200	+2.4800	3.2000	9.7440	32.2560	76.8%
0.2	nearly sticks	+0.2400	+1.8400	1.6000	5.1360	36.8640	87.8%
0.0	perfectly inelastic — they move off together	+1.2000	+1.2000	0.0000	3.6000	38.4000	91.4%

Two equations, one parameter, and the four results every collision question wants

Every head-on collision between two bodies is the same problem with one dial on it. You have two unknowns, the two final velocities, so you need two equations. The first is conservation of momentum, which holds always. The second is Newton’s experimental law: the relative speed of separation is e times the relative speed of approach. That is all. The elastic formulae and the perfectly inelastic formula that textbooks present as separate results are the e = 1 and e = 0 cases of the same pair of equations, and keeping them separate in your head is a good way to memorise four things instead of understanding one.

e is a ratio of RELATIVE speeds, which is the definition people misremember. It is not the ratio of one body’s speed before and after, except in the special case where the other body does not move — which is exactly the case of a ball bouncing off the floor, and is why the misremembering survives. For a ball dropped from rest onto a hard surface, e = √(h_rebound/h_drop), and that is the standard school measurement of it. For two bodies that both move, go back to the definition.

Momentum is conserved whether or not energy is, and the check costs one line. This page prints the total momentum before and after and they must agree to the last figure printed, at every e. Kinetic energy, by contrast, is conserved only when e = 1. At e = 0 the loss is as large as it can be — but note that it is not everything. A perfectly inelastic collision keeps the kinetic energy of the combined body’s motion, which momentum conservation will not allow it to surrender; in the worked example that is 3.6 J out of the original 42. The only collisions that lose all of their kinetic energy are the ones whose total momentum was zero to begin with.

The centre of mass is the quiet one. No external force acts during the collision, so the total momentum does not change, so the velocity of the centre of mass does not change either — before, during and after, at every value of e. The page prints it. Once you have seen that, the role of e becomes clear: it decides how fast the two bodies move relative to the centre of mass after the impact, as a fraction of how fast they were moving relative to it before. At e = 1 that fraction is 1 and nothing is lost; at e = 0 it is zero and they travel with the centre of mass. The CBSE Class XI treatment puts centre of mass and momentum conservation side by side in its rigid-body unit for this reason.

Which boards want e — and the surprise is how few. The coefficient of restitution does not appear anywhere in the CBSE Class XI physics list: Unit IV covers elastic and inelastic collisions in one and two dimensions without naming e. Cambridge 9702 goes further and says outright that “knowledge of the concept of coefficient of restitution is not required”, while still requiring that in an elastic collision the relative speed of approach equals the relative speed of separation — which is e = 1 described rather than named. Where it IS examined is mathematics: Pearson’s Further Mathematics specification lists “Direct impact of elastic spheres. Newton’s law of restitution” in Further Mechanics 1, with 0 ≤ e ≤ 1 explicitly examinable, and goes on to successive and oblique impacts. So if you are meeting e for the first time in a physics lesson, check which paper it is for.

Frequently asked questions

What is the coefficient of restitution?

The relative speed at which two bodies separate divided by the relative speed at which they approached: e = (v₂ − v₁)/(u₁ − u₂). It runs from 0 to 1. At 1 the collision is perfectly elastic and the relative velocity simply reverses; at 0 the bodies do not separate at all and move off together. It is a property of the two materials and, less conveniently, of the impact speed, so a value measured at one speed is only a guide at another.

How do I measure e for a ball?

Drop it from rest onto a hard, heavy, flat surface and measure the rebound height. Then e = √(h_rebound/h_drop), because the speeds at impact and rebound go as the square roots of the heights. A ball returning to half its drop height has e = √0.5 = 0.71. This works because the floor does not move, so the relative speeds are just the ball’s speeds; for two bodies that both move you must use the full definition.

What is conserved in an inelastic collision?

Momentum, always. Kinetic energy, no. That asymmetry is the key to the whole topic: write the momentum equation with confidence in every collision problem, and only add an energy equation when you have been told the collision is elastic. Total energy is conserved, of course — the missing kinetic energy has gone into deformation, heating and sound — but it is no longer available as motion and no longer appears in your equations.

Does a perfectly inelastic collision lose all the kinetic energy?

No, and this is worth being clear about. It loses the maximum that conservation of momentum permits, which still leaves the kinetic energy of the combined body moving at the velocity of the centre of mass. In the worked example above, 38.4 J of 42 J goes and 3.6 J remains. Only a collision in which the total momentum is zero — two identical bodies meeting head-on at equal speeds — can lose the whole of it, and then both bodies end up at rest.

Two equal masses collide elastically and one was at rest. What happens?

They exchange velocities: the moving one stops dead and the stationary one leaves with the whole of the incoming velocity. Put m₁ = m₂, u₂ = 0 and e = 1 into this page and you will see it. It is the result behind a snooker break, behind a Newton’s cradle, and behind the use of light nuclei as neutron moderators — a neutron loses the most energy to a nucleus of its own mass, which is why hydrogen moderates better than lead.

Why do the final velocities still come out when the bodies are not even approaching?

Because the algebra does not know that. Feed it two bodies that are separating and it will solve the same two equations and hand you numbers; they describe a collision that is not going to happen. The page says so in a note when the approach speed is zero or negative. The usual cause is a sign: both velocities must be measured along the same positive direction, and a head-on collision needs them to have opposite signs.

Related calculators

References

Pearson Edexcel Level 3 Advanced GCE in Further Mathematics (9FM0), specification Issue 4, June 2023 (read 11 November 2026). Further Mechanics 1 carries what the physics syllabuses leave out: topic 1.1 is “Momentum and impulse. The impulse-momentum principle”, topic 4.1 is “Direct impact of elastic spheres. Newton’s law of restitution”, with the note that “Students will be expected to know and use the inequalities 0 ≤ e ≤ 1”, and topics 4.2 and 5 extend to successive and oblique impacts. So the coefficient of restitution is examinable — in mechanics, under mathematics, on a different paper from the physics.
Cambridge International AS & A Level Physics 9702, syllabus for examination in 2025, 2026 and 2027, Cambridge Assessment International Education (read 11 November 2026). Two exclusions on this syllabus are worth more than most inclusions, because they explain why a question from one board looks unanswerable to a student taught on another. Topic 3.2 asks only for “a qualitative understanding of frictional forces and viscous/drag forces including air resistance” and states that “no treatment of the coefficients of friction and viscosity is required”. Topic 3.3 states that “knowledge of the concept of coefficient of restitution is not required”, while still requiring that for an elastic collision “the relative speed of approach is equal to the relative speed of separation”. Its data sheet gives the acceleration of free fall as 9.81 m s⁻².
PHYSICS (042) Curriculum, Class XI–XII, Central Board of Secondary Education, session 2025–26 (read 11 November 2026). Unit III, Laws of Motion, reads in part: “Intuitive concept of force, Inertia, Newton’s first law of motion; momentum and Newton’s second law of motion; impulse”, “Equilibrium of concurrent forces, Static and kinetic friction, laws of friction, rolling friction, lubrication” and “Dynamics of uniform circular motion: Centripetal force, examples of circular motion (vehicle on a level circular road, vehicle on a banked road)”. Unit V, Motion of System of Particles and Rigid Body, reads in part: “Moment of a force, torque, angular momentum, law of conservation of angular momentum and its applications” and “Moment of inertia, radius of gyration, values of moments of inertia for simple geometrical objects (no derivation)”. Quoted to establish what is in scope, not to claim any endorsement by the board.
The International System of Units (SI), 9th edition, 2019, Bureau International des Poids et Mesures (read 11 November 2026). Table 4 gives the newton as kg m s⁻², which is the only definition of force this site uses: a newton is not a unit of mass and a kilogram is not a unit of force. Appendix 1 lists the 3rd CGPM of 1901, whose declaration both separated mass from weight and fixed the conventional value of g_n — the two things a force question most often conflates. Cited, not reproduced.
Fundamental Physical Constants — Complete Listing, NIST Reference on Constants, Units and Uncertainty, 2022 CODATA adjustment (read 11 November 2026). The line this batch depends on is “standard acceleration of gravity 9.806 65 (exact) m s⁻²”. The word exact is the point: g_n is a conventional value, not a measured one, so it carries no uncertainty and it is not the g where you are standing. The same listing gives the Newtonian constant of gravitation as 6.674 30 × 10⁻¹¹ m³ kg⁻¹ s⁻² with an uncertainty of 0.000 15 × 10⁻¹¹, which is 22 parts per million — the worst-known of the constants a school paper quotes, and a useful reminder that “constant” and “exact” are different claims.